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## Self Balancing Binary Search Tree Multiple choice Questions and Answers (MCQs)

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Question 1 |

Which of the following is not the self balancing binary search tree?

AVL Tree | |

2-3-4 Tree | |

Red - Black Tree | |

Splay Tree |

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Question 1 Explanation:

2-3-4 Tree is balanced search trees. But it is not a binary tree. So, it is not a self balancing binary tree. AVL tree, Red-Black Tree and Splay tree are self balancing binary search tree.

Question 2 |

The binary tree sort implemented using a self - balancing binary search tree takes ..... time is worst case.

O(n log n) | |

O(n) | |

O(n ^{2}) | |

O(log n) |

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Question 2 Explanation:

The worst case running time of the binary tree sort is O(n

^{2}). But, the worst case running time can be improved to the O(n log n) using a self - balancing binary search tree.

Question 3 |

An AVL tree is a self - balancing binary search tree, in which the heights of the two child sub trees of any node differ by .....

At least one | |

At most one | |

Two | |

At most two |

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Question 3 Explanation:

In an AVL tree, the difference between heights of the two child sub trees of any node is at most one. If the height differs by more than one, AVL tree performs rotations to balance the tree.

Question 4 |

Associative arrays can be implemented using .....

B-tree | |

A doubly linked list | |

A single linked list | |

A self balancing binary search tree |

**Puzzles Questions answers**

Question 4 Explanation:

Associative arrays can be implemented using a self balancing binary search tree as the worst-case time performance of self - balancing binary search trees is O(log n).

Question 5 |

Self - balancing binary search trees have a much better average-case time complexity than hash tables.

True | |

False |

**NTA NET Questions answers**

Question 5 Explanation:

For lookup, insertion and deletion hash table take O(1) time in average-case while self - balancing binary search trees takes O(log n). Therefore, hash tables perform better in average-case.

There are 5 questions to complete.